5 Bomb testing

This time the setup is a Mach−Zehnder interferometer, which consists of two beam splitters (S1 and S2), two mirrors (M1 and M2), and two photodetectors (D1 and D2) arranged as in Fig. 1.4.1. A particular twist of the experiment we are about to discuss[1] is the possible presence of a “bomb” — a photodetector so sensitive that it will explode if it absorbs a single photon. (For simplicity’s sake we make the usual assumption that all detectors including the bomb are 100% efficient.)

Figure 1.5.1 The “Bomb testing” experiment of Elitzur and Vaidman

Imagine, to begin with, that neither S2 nor the bomb is present. A beam of photons (a.k.a. a light beam) enters S1 from the left. Classically described, two beams emerge, each with half the intensity of the incoming beam. Described in quantum-mechanical terms, each incoming photon has a 50% chance of being detected by D1 (indicating that the photon was reflected upward by S1) and an equal chance of being detected by D2 (indicating that the photon went horizontally through S1).

If S2 (but as yet no bomb) is present, Rule B applies. Here is what we need to know about the amplitudes associated with the alternatives (reflection by M1 or reflection by M2): they are equal except that each reflection causes a phase shift of 90°. In other words, it rotates the amplitude anticlockwise by 90°. This is equivalent to multiplying it by i = √(−1). (The magnitude of the phase shift depends on the materials used. For the sake of convenience we imagine using materials for which it equals 90°.)

Each of the alternatives leading to D1 involves two reflections, so the corresponding amplitudes are equal (i2A, say). The probability of detection by D1 is therefore given by

pB,1 = |i2A + i2A|2 = 4|A|2.

The alternative leading to D2 via M1 involves three reflections, so the corresponding amplitude equals i3A, while the alternative leading to D2 via M2 involves a single reflection, the corresponding amplitude thus being iA. Since the two amplitudes differ by a factor i2 = −1, the probability of detection by D2 is

pB,2 = |iA + i3A|2 = |iA − iA|2 = 0.

Finally, if both S2 and the bomb are present, the alternative taken by the photon is measured. If the bomb explodes, this indicates that the photon went via M1, and if it does not explode, this indicates that the photon went via M2. If it went via M2, either photodetector responds with probability 1/2. Thus:

If the bomb is absent, D1 “clicks” every time (in 100% of all cases), whereas D2never “clicks”.

If the bomb is present, it explodes half of the time (in 50% of all cases); and if it doesn’t explode, D1 and D2 are equally likely to respond (each “clicks” in 25% of all cases).

Now suppose that the bomb is present. Is it possible, using the present setup, to ascertain the presence of the bomb without setting it off? Stop to think about this before you proceed.

The answer is affirmative, albeit only in 25% of the tests. If the bomb explodes, which happens in 50% of the tests, we have failed. If the bomb is present and D1 responds, which happens in 25% of the tests, we have learned nothing, for D1 also responds if the bomb is absent. But if D2 responds, which happens in the remaining 25% of the tests, we have succeeded, for D2 would not have responded if the bomb had been absent.

When a version of this experiment was demonstrated at a science fair in Groningen, the Netherlands, in 1995, the reactions of non-physicists differed markedly from those of physicists.[2] Everyone was perplexed, for the detection of the photon by D2 seems to have contradictory implications:

The bomb was present.

The photon never came near the bomb.

If the photon never came near the bomb, how was it possible to learn that the bomb was present? While most ordinary folks thought that some physicist will eventually solve this puzzle, the physicists themselves were decidedly less hopeful that a satisfactory explanation will be found.